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Question Number 44302 by abdo.msup.com last updated on 26/Sep/18

calculate ∫_0 ^∞ (dx/((x+1)(x+2)(x+3)))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)} \\ $$

Commented by maxmathsup by imad last updated on 28/Sep/18

let I =∫_0 ^∞ (dx/((x+1)(x+2)(x+3))) let decompose F(x)=(1/((x+1)(x+2)(x+3))) F(x)=(a/(x+1)) +(b/(x+2)) +(c/(x+3)) a =lim_(x→−1) (x+1)F(x)=(1/2) b=lim_(x→−2) (x+2)F(x) = −1 c =lim_(x→−3) (x+3)F(x)=(1/2) ⇒F(x)= (1/(2(x+1))) −(1/((x+2))) +(1/(2(x+3))) ⇒∫_0 ^∞ F(x)dx =∫_0 ^∞ { (1/(2(x+1))) −(1/(x+2)) +(1/(2(x+3)))}dx =[(1/2)ln∣x+1∣−ln∣x+2∣+(1/2)ln∣x+3∣]_0 ^(+∞) =[ln(((√((x+1)(x+3)))/(x+2))]_0 ^(+∞) −ln(((√3)/2)) =ln(2)−(1/2)ln(3) ⇒ I =ln(2)−((ln(3))/2) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{{x}+\mathrm{2}}\:+\frac{{c}}{{x}+\mathrm{3}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${b}={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=\:−\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{3}} \left({x}+\mathrm{3}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow{F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{3}\right)} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{F}\left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\infty} \:\left\{\:\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{{x}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{3}\right)}\right\}{dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}+\mathrm{1}\mid−{ln}\mid{x}+\mathrm{2}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}+\mathrm{3}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\left[{ln}\left(\frac{\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)}}{{x}+\mathrm{2}}\right]_{\mathrm{0}} ^{+\infty} \right. \\ $$$$−{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)\:\Rightarrow\:{I}\:={ln}\left(\mathrm{2}\right)−\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{2}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18

(1/((x+1)(x+2)(x+3)))=(a/(x+1))+(b/(x+2))+(c/(x+3)) 1=a(x+2)(x+3)+b(x+1)(x+3)+c(x+1)(x+2) put x+1=0 x=−1 1=a(−1+2)(−1+3) a=(1/2) x+2=0 1=b(−2+1)(−2+3) b=−1 x+3=0 1=c(−3+1)(−3+2) c=(1/2) ∫_0 ^∞ ((1/2)/(x+1))dx+∫_0 ^∞ ((−1)/(x+2))dx+∫_0 ^∞ ((1/2)/(x+3))dx {(1/2)ln∣x+1∣−ln∣x+2∣+(1/2)ln∣x+3∣}_0 ^∞ ={ln∣((√((x+1)(x+3)))/(x+2))∣}_0 ^∞ ={ln∣((x.(√((1+(1/x))(1+(3/x)))))/(x(1+(2/x))))∣}_0 ^∞ ={ln1−ln∣((√(1×3))/2)∣} =0−{(1/2)ln3−ln2} =ln2−(1/2)ln3

$$\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}=\frac{{a}}{{x}+\mathrm{1}}+\frac{{b}}{{x}+\mathrm{2}}+\frac{{c}}{{x}+\mathrm{3}} \\ $$$$\mathrm{1}={a}\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+{b}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)+{c}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right) \\ $$$${put}\:{x}+\mathrm{1}=\mathrm{0}\:\:\:{x}=−\mathrm{1} \\ $$$$\mathrm{1}={a}\left(−\mathrm{1}+\mathrm{2}\right)\left(−\mathrm{1}+\mathrm{3}\right)\:\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{1}={b}\left(−\mathrm{2}+\mathrm{1}\right)\left(−\mathrm{2}+\mathrm{3}\right)\:\:{b}=−\mathrm{1} \\ $$$${x}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{1}={c}\left(−\mathrm{3}+\mathrm{1}\right)\left(−\mathrm{3}+\mathrm{2}\right)\:\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{2}}}{{x}+\mathrm{1}}{dx}+\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{1}}{{x}+\mathrm{2}}{dx}+\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{2}}}{{x}+\mathrm{3}}{dx} \\ $$$$\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}+\mathrm{1}\mid−{ln}\mid{x}+\mathrm{2}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}+\mathrm{3}\mid\right\}_{\mathrm{0}} ^{\infty} \\ $$$$=\left\{{ln}\mid\frac{\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)}}{{x}+\mathrm{2}}\mid\right\}_{\mathrm{0}} ^{\infty} \\ $$$$=\left\{{ln}\mid\frac{{x}.\sqrt{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)}}{{x}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}\mid\right\}_{\mathrm{0}} ^{\infty} \\ $$$$=\left\{{ln}\mathrm{1}−{ln}\mid\frac{\sqrt{\mathrm{1}×\mathrm{3}}}{\mathrm{2}}\mid\right\} \\ $$$$=\mathrm{0}−\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}−{ln}\mathrm{2}\right\} \\ $$$$={ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3} \\ $$

Commented by maxmathsup by imad last updated on 28/Sep/18

your answer is correct sir Tanmay thanks...

$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Tanmay}\:{thanks}... \\ $$

Answered by ajfour last updated on 28/Sep/18

let x+2 = z ⇒I=∫_2 ^( ∞) (dz/(z(z^2 −1))) = (1/2)∫(dz/(z+1))−∫ (dz/z)+(1/2)∫(dz/(z−1)) = (1/2)ln ∣((z^2 −1)/z^2 )∣_2 ^∞ = (1/2)ln (4/3) = ln 2−(1/2)ln 3 .

$${let}\:\:{x}+\mathrm{2}\:=\:{z} \\ $$$$\Rightarrow{I}=\int_{\mathrm{2}} ^{\:\:\infty} \frac{{dz}}{{z}\left({z}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dz}}{{z}+\mathrm{1}}−\int\:\frac{{dz}}{{z}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dz}}{{z}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{2}} }\mid_{\mathrm{2}} ^{\infty} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{3}}\:=\:\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}\:. \\ $$

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